Here is Prob. 14, Sec. 30, in the book Topology by James R. Munkres, 2nd edition:
Show that if $X$ is Lindelof and $Y$ is compact, then then $X \times Y$ is Lindelof.
My Attempt:
Let $X$ and $Y$ be topological spaces such that $X$ is Lindelof and $Y$ is compact.
Now let $\mathscr{A}$ be an open covering of the product space $X \times Y$.
Let $x \in X$. Then let $\mathscr{A}_x$ be the subcollection of $\mathscr{A}$ such that $\mathscr{A}_x$ covers $\{x\} \times Y$, which is compact being homeomorphic with the compact space $Y$. Therefore some finite subcollection of $\mathscr{A}_x$ also covers $\{x \} \times Y$; let one such subcollection be
$$
\left\{ \mathbf{A}_{x, 1}, \ldots, \mathbf{A}_{x, n_x} \right\}, \tag{Definition 0}
$$
and then let us put
$$
\mathbf{S}_x \colon= \bigcup_{k=1}^{n_x} \mathbf{A}_{x, k}. \tag{Definition 1}
$$
Then $\mathbf{S}_x$ is an open set of $X \times Y$ such that $\mathbf{S}_x$ covers $\{x \} \times Y$, and since $Y$ is compact, therefore by Lemma 26.8 (The tube lemma) in Munkres there exists an open set $U_x$ of $X$ such that $x \in U_x$ and such that $\mathbf{S}_x$ also covers the tube $U_x \times Y$.In this way we obtain an open covering $\left\{ U_x \colon x \in X \right\}$ for the Lindelof space $X$, and so a countable subcollection $\left\{ U_{x_1}, U_{x_2}, U_{x_3}, \ldots \right\}$ of the collection $\left\{ U_x \colon x \in X \right\}$ also covers $X$. Thus the collection
$$ \left\{ \mathbf{S}_{x_1}, \mathbf{S}_{x_2}, \mathbf{S}_{x_3}, \ldots \right\} $$
covers $X \times Y$.Therefore using (Definition 0) and (Definition 1) above we can conclude that the collection
$$
\bigcup_{j=1}^\infty \left\{ \mathbf{A}_{x_j, k} \colon k = 1, \ldots, n_{x_j} \right\}
$$
is a countable subcollection of $\mathscr{A}$ that covers $X \times Y$.Thus every open covering $\mathscr{A}$ of $X \times Y$ has a countable subcollection that also covers $X \times Y$.
Hence $X \times Y$ is Lindelof.
Is this proof correct and accurate in each and every detail? Or, are there any errors in my reasoning?
Best Answer
Yes, the proof works and is a copy-paste of 26.7 , the proof that the product of $X \times Y$ is compact when $X,Y$ are (except that a finite subcover must be replaced by a countable one). From that product proof Munkres abstracts the tube lemma that you re-use here for a Lindelöf factor. The write-up I have no issues with.
Note that the product of two Lindelöf spaces is not always Lindelöf, as shown by $\Bbb R_l \times \Bbb R_l$, the "canonical" example of this phenomenon, so the compact factor is quite essential.
There is another generalisation to be made here (due to Michael, I believe, and covered in Engelking's excellent text book General Topology (2nd ed.)): if $f: X \to Y$ is a perfect map (onto, continuous, closed and all fibres $f^{-1}[\{y\}], y \in Y$ are compact) and $Y$ is Lindelöf, then so is $X$. This is a generalisation, because if $X$ is compact, the projection map $\pi_Y: X \times Y \to Y$ is closed by that same tube lemma; the other parts of the definition of a perfect map for $\pi_Y$ are clear, so from that generalised theorem we can also conclude from the Lindelöfness of the non-compact factor that the whole product is Lindelöf.